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[ewald] dudopráce 

Former-commit-id: bd5a14007237ffcc3dbb0110da0b68914916c980
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Marek Nečada 2017-08-15 02:29:58 +00:00
parent 83fed81e24
commit 23d1248708
1 changed files with 43 additions and 4 deletions
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notes/ewald.lyx
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@ -187,6 +187,11 @@
\end_inset \end_inset
\begin_inset FormulaMacro
\newcommand{\hgf}[4]{\mathbf{F}\left(#1,#2;#3;#4\right)}
\end_inset
\end_layout \end_layout
\begin_layout Title \begin_layout Title
@ -876,6 +881,40 @@ leads to satisfactory results, as will be shown below.
Hankel transforms of the long-range parts Hankel transforms of the long-range parts
\end_layout \end_layout
\begin_layout Standard
From REF DLMF 10.22.49
\size footnotesize
\begin_inset Formula
\begin{eqnarray*}
\Xi_{c-ik_{0}}^{q,n}(k) & \equiv & \int_{0}^{\infty}e^{-cr}e^{ik_{0}r}\left(k_{0}r\right)^{-q}J_{n}\left(kr\right)r\,\ud r\\
& = & k_{0}^{-q}\int_{0}^{\infty}r^{2-q-1}e^{-(c-ik_{0})r}J_{n}(br)\,\ud r\\
& = & k_{0}^{-q}\frac{\left(\frac{k}{2}\right)^{n}}{\left(c-ik_{0}\right)^{2-q+n}}\Gamma\left(2-q+n\right)\hgf{\frac{2-q+n}{2}}{\frac{3-q+n}{2}}{n+1}{-\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}\\
& & \quad\Re\left(2-q+n\right)>0,\Re\left(c-ik_{0}\pm k\right)>0.
\end{eqnarray*}
\end_inset
\size default
This by itself does not provide too much insight, but fortunately the hypergeome
tric function has more comprehensive expressions for special arguments (this
is from Mathematica):
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\Xi_{c-ik_{0}}^{1,n}(k) & = & k_{0}^{-1}k^{n}\frac{\left(1+\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}\right)^{-n}\left(c-ik_{0}\right)^{-1-n}}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\\
\Xi_{c-ik_{0}}^{2,n}(k) & = & k_{0}^{-2}k^{n}\frac{\left(1+\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}\right)^{-n}\left(c-ik_{0}\right)^{-n}}{n},\quad n>0\\
\Xi_{c-ik_{0}}^{2,n}(k) & = & k_{0}^{-3}k^{n}\frac{\left(1+\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}\right)^{-n}}{n(n^{2}-1)}
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection \begin_layout Subsection
3d 3d
\end_layout \end_layout
@ -941,7 +980,7 @@ where the spherical Hankel transform
2) 2)
\begin_inset Formula \begin_inset Formula
\[ \[
\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right). \bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
\] \]
\end_inset \end_inset
@ -951,7 +990,7 @@ Using this convention, the inverse spherical Hankel transform is given by
3) 3)
\begin_inset Formula \begin_inset Formula
\[ \[
g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k), g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
\] \]
\end_inset \end_inset
@ -964,7 +1003,7 @@ so it is not unitary.
An unitary convention would look like this: An unitary convention would look like this:
\begin_inset Formula \begin_inset Formula
\begin{equation} \begin{equation}
\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition} \usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
\end{equation} \end{equation}
\end_inset \end_inset
@ -1018,7 +1057,7 @@ where the Hankel transform of order
is defined as is defined as
\begin_inset Formula \begin_inset Formula
\begin{equation} \begin{equation}
\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition} \pht mg\left(k\right)=\int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
\end{equation} \end{equation}
\end_inset \end_inset